details

property	value
status	complete
benchmark	Velroyen08-twoFloatInterv.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	39.0207929611 seconds
cpu usage	41.736331773
max memory	2.9138944E7

stage attributes

key	value
output-size	2721
starexec-result	MAYBE

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] f3#(I3, I4) -> f2#(I3 + 1, I5) [39 <= I3 - 1] f3#(I6, I7) -> f2#(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] f3#(I9, I10) -> f2#(I9, I11) [I9 <= 10 /\ I9 <= 29] f2#(I12, I13) -> f3#(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] f2#(I15, I16) -> f3#(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] f1#(I18, I19) -> f2#(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] f3(I3, I4) -> f2(I3 + 1, I5) [39 <= I3 - 1] f3(I6, I7) -> f2(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] f3(I9, I10) -> f2(I9, I11) [I9 <= 10 /\ I9 <= 29] f2(I12, I13) -> f3(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] f2(I15, I16) -> f3(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] f1(I18, I19) -> f2(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] The dependency graph for this problem is: 0 -> 7 1 -> 6 2 -> 6 3 -> 5, 6 4 -> 5 5 -> 3, 4 6 -> 1, 2, 3 7 -> 5, 6 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] 2) f3#(I3, I4) -> f2#(I3 + 1, I5) [39 <= I3 - 1] 3) f3#(I6, I7) -> f2#(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] 4) f3#(I9, I10) -> f2#(I9, I11) [I9 <= 10 /\ I9 <= 29] 5) f2#(I12, I13) -> f3#(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] 6) f2#(I15, I16) -> f3#(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] 7) f1#(I18, I19) -> f2#(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] We have the following SCCs. { 1, 2, 3, 4, 5, 6 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] f3#(I3, I4) -> f2#(I3 + 1, I5) [39 <= I3 - 1] f3#(I6, I7) -> f2#(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] f3#(I9, I10) -> f2#(I9, I11) [I9 <= 10 /\ I9 <= 29] f2#(I12, I13) -> f3#(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] f2#(I15, I16) -> f3#(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] f3(I3, I4) -> f2(I3 + 1, I5) [39 <= I3 - 1] f3(I6, I7) -> f2(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] f3(I9, I10) -> f2(I9, I11) [I9 <= 10 /\ I9 <= 29] f2(I12, I13) -> f3(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] f2(I15, I16) -> f3(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] f1(I18, I19) -> f2(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] popout

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